Two masses of an Atwood machine are at rest on a tabletop. A small upward force | TestHub

PhysicsEasymultiple choice

Two masses of an Atwood machine are at rest on a tabletop. A small upward force $F_{p}$ on the pulley keeps the string taut. Suppose that $m_{2} > m_{1}$ . Assume that the pulley itself has negligible mass. Mark the CORRECT statement(s) :-

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Answer:

A, B, C

Solution:

F.B.D of pulley

$2 T = F_{P}$

$T = F_{p} /2$

F.B.D. of $m_{1} : T + N = m_{1} g$

for block to leave the table, N must be zero.

$T = m_{1} g$

$\frac{F _{P}}{2} = m_{1} g = F_{P} = 2 m_{1} g$

so for $∣ F_{p} ∣ < 2 m_{1} ∣ g ∣ : m_{1}$ remains at rest \& since $m_{2} > m_{1}, m_{2}$ also remains at rest. Now for block of mass $m_{2}$ :

To lift off the table, N should be zero.

$T + N = m_{2} g = T = m_{2} g$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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